// Numbas version: exam_results_page_options {"name": "Ida Friestad's copy of Integration: Indefinite integral by substitution", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"extensions": [], "advice": "

This exercise is best solved by using substitution.

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Note that the numerator $\\simplify[std]{{2 * a} * x + {b}}$ of \\[\\simplify[std]{({2 * a} * x + {b}) / ({a} * x ^ 2 + {b} * x + {c})}\\] is the derivative of the denominator $\\simplify[std]{{a} * x ^ 2 + {b} * x + {c}}$

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So if you use as your substitution $u=\\simplify[std]{{a} * (x ^ 2) + ({b} * x) + {c}}$ you then have $\\simplify[std]{ du = ({2 * a} * x + {b}) * dx}$

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Hence we can replace $\\simplify[std]{ ({2 * a} * x + {b}) * dx}$ by $du$

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Hence the integral becomes:

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\\[\\begin{eqnarray*} I&=&\\int\\;\\frac{du}{u}\\\\ &=&\\ln(|u|)+C\\\\ &=& \\simplify[std]{ln(abs({a} * (x ^ 2) + ({b} * x) + {c}))+C} \\end{eqnarray*}\\]

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A Useful Result
This example can be generalised.
Suppose \\[I = \\int\\; \\frac{f'(x)}{f(x)}\\;dx\\]
The using the substitution $u=f(x)$ we find that $du=f'(x)\\;dx$ and so using the same method as above:
\\[I = \\int \\frac{du}{u} = \\ln(|u|)+ C = \\ln(|f(x)|)+C\\]

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Find the following integral.

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Input the constant of integration as $C$.

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Input all numbers as integers or fractions.

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Do not input numbers as decimals, only as integers without the decimal point, or fractions

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Try the substitution $u=\\simplify[std]{{a} * (x ^ 2) + ({b} * x) + {c}}$

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\\[I=\\simplify[std]{Int(({2 * a} * x + {b}) / ({a} * x ^ 2 + {b} * x + {c}),x)}\\]

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$I=\\;$[[0]]

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Input the constant of integration as $C$.

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Input all numbers as integers or fractions not as decimals.

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Click on Show steps if you need help. You will lose 1 mark if you do so.

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Find $\\displaystyle \\int \\frac{2ax + b}{ax ^ 2 + bx + c}\\;dx$

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